Fun squared

First I'll show only what's necessary:

From AM-GM inequality, we have $\begin{aligned} x_1^2+\frac{1}{3}x_2^2 &\ge \frac{2}{\sqrt{3}}x_1x_2\\ \frac{2}{3}x_2^2+\frac{1}{2}x_3^2 &\ge \frac{2}{\sqrt{3}}x_2x_3\\ \frac{1}{2}x_3^2+\frac{2}{3}x_4^2 &\ge \frac{2}{\sqrt{3}}x_3x_4\\ \frac{1}{3}x_4^2+x_5^2 &\ge \frac{2}{\sqrt{3}}x_4x_5\\ \end{aligned}$ Sum them up, we get $x_1^2+x_2^2+x_3^2+x_4^2+x_5^2 \ge \frac{2}{\sqrt{3}}(x_1x_2+x_2x_3+x_3x_4+x_4x_5)$ It follows immediately that the maximum of $S$ is $\boxed{3}$ . It occurs when $(x_1,x_2,x_3,x_4,x_5) = \left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{2}},\sqrt{\frac{2}{3}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{6}}\right).$

Here's how I know why I should divide them like that:

We first notice that the summation inside $S$ is not cyclic. So we need to come up with a new kind of separation. Let $0<k<1$ be an unknown real number. Since we have $\sum x_i^2=2$ , we should do something with it. We start out by $x_1^2+kx_2^2 \ge 2\sqrt{k}x_1x_2.$ The number $kx_2^2$ is a fraction of $x_2^2$ that we want to combine with $x_1^2$ . It means that we use only $kx_2^2$ , not an entire $x_2^2$ . Now we have $(1-k)x_2^2$ left. Our next goal is to combine with a fraction of $x_3^2$ . The question arises that how much do I need to grab from an entire $x_3^2$ ? The answer is $\frac{k}{1-k}$ , why? Because we need a fraction of $x_3^2$ so that when we apply AM-GM, the right hand side is still $2\sqrt{k}x_2x_3$ , just like the first one. If we do like that, when we sum up, we would get $2\sqrt{k} \sum x_ix_{i+1}$ . Now, we have $(1-k)x_2^2+\left(\frac{k}{1-k}\right)x_3^2 \ge 2\sqrt{k}x_2x_3.$ Similarly, we now have $\left(1-\frac{k}{1-k}\right)x_3^2=\left(\frac{1-2k}{1-k}\right)x_3^2$ left. Iterate this method, we will get these inequalities:- $\left(\frac{1-2k}{1-k}\right)x_3^2 +\left(\frac{k-k^2}{1-2k}\right)x_4^2 \ge 2\sqrt{k}x_3x_4,$ $\left(\frac{1-3k+k^2}{1-2k}\right)x_4^2+ \left(\frac{k-2k^2}{1-3k+k^2}\right)x_5^2 \ge 2\sqrt{k}x_4x_5.$ Now, the right hand side is ready, but we still have a fraction of $x_5^2$ left, what should we do? The solution is take it all. Meaning that we choose $k$ such that the fraction of $x_5^2$ in the last inequality is actually an entire $x_5^2$ . So we set $\frac{k-2k^2}{1-3k+k^2}=1.$ After solving this equation, we have $k=1/3$ that is between zero and one. All we need to do after this is to check that it really works just like what I wrote above. That's the trick for question like this.

Here is a solution for those who are familiar with quadratic forms .

The symmetric matric of the quadratic form $q(x_1,...,x_5)=2\sum_{k=1}^{4}x_kx_{k+1}$ is $A=\begin{bmatrix}0&1&0&0&0\\1&0&1&0&0\\0&1&0&1&0\\0&0&1&0&1\\0&0&0&1&0\end{bmatrix}$ The characteristic polynomial is $\det(A-xI_5)=-x(x^2-1)(x^2-3)$ , so that the largest eigenvalue of $A$ is $\sqrt{3}$ . This is the maximal value of $\sum_{k=1}^{4}x_kx_{k+1}$ when $\sum_{k=1}^{5}x_k^2=2$ , with $S=\boxed{3}$ .

The idea behind this technique is the following: Once we know the eigenvalues $\pm\sqrt{3},\pm 1,0$ of $q$ , we can perform a change of variables and write the form as $\sqrt{3}v_1^2+v_2^2-v_3^2-\sqrt{3}v_4^2$ . To maximize the value of $q$ when $\sum_{k=1}^{5}v_k^2=2$ we make $v_1^2=2$ and $v_k=0$ for $k>1$ to obtain $2\sqrt{3}$ . The sum we seek to maximize is $\frac{q}{2}$ .

(Equivalently, this solution can be phrased in the language of Lagrange multipliers.)